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Licemer1 [7]
4 years ago
12

A $20 bill two $10 bills three $5 bills and four $1 bills are placed in a bag. If a bill is chosen at random what is the expecte

d value for the amount chosen?
Mathematics
1 answer:
muminat4 years ago
8 0

Answer: Hello mate!

You have one $20 bill, two $10 bills, three $5 bills and four $1 bills in a bag.

This adds to a total of 10 bills in the bag, and you want to know the expected value if you took one at random.

The expected value is defined as ∑pₙxₙ

where pₙ is the probability for each event (in this case is the number of bills of a given price divided the number of bills in the bag) and xₙ is the value of the bill.

Then we can write the expected value as:

E = (1/10)*$20 + (2/10)*$10 + (3/10)*$5 + (4/10)*$1 = (1/10)( 20$ + 2*$10 + 3*$5 + 4*$1) = (1/10)*($59) = $5.90

The expected value is $5.90.

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Answer:

Step-by-step explanation:

Given:

Cost of L = $5

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Solution:

L + S = 60 - (1) (multiplying by 3)

5L + 3S = 220 - (2)

(-)3L +(-) 3S = (-)180 - (1)

2L = 40

L = 40 / 2

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Answer:

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Step-by-step explanation:

Given:

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r=\sqrt[3]{\frac{3V}{4\pi }}

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